The figure show a rod AB, which is bent in a 120° circular arc of radius R. A charge (-Q) is uniformly distributed over rod AB. What is the electric field \(\vec{E}\) at the centre of curvature O?
(1) \(\frac{3 \sqrt 3 Q}{8 \pi \varepsilon_0 R^2} (\hat {i})\)
(2) \(\frac{3 \sqrt 3 Q}{8 \pi^2 \varepsilon_0 R^2} (\hat {i})\)
(3) \(\frac{3 \sqrt 3 Q}{16 \pi^2 \varepsilon_0 R^2} (\hat {i})\)
(4) \(\frac{3 \sqrt 3 Q}{8 \pi^2 \varepsilon_0 R^2} (-\hat{i})\)
Correct answer: (2) \(\frac{3 \sqrt 3 Q}{8 \pi^2 \varepsilon_0 R^2} (\hat {i})\)
Explanation:
ε = \(\frac{2k \lambda}{R}sin\)\(\frac{\theta}{2}(\hat{i})\)
λ = \(\Big(\frac{-Q}{R \theta}\Big)\) = \(\Big(\frac{-Q}{R. \frac{2 \pi}{3}}\Big)\)
λ = \(\frac{-3Q}{2 \pi R}\)
ε = \(\frac{2k}{R}\).\(\frac{-3Q}{2 πR}sin60°(-\hat {i})\)
ε = \(\frac{3 \sqrt 3 Q}{8 \pi^2 \varepsilon_0 R^2} ( \hat {i})\)